1. Monday, Apr. 7, 2008 PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 1 PHYS 1441 – Section 002 Lecture #19 Monday, Apr. 7, 2008 Dr. Jaehoon Yu Linear Momentum Conservation Collisions Center of Mass Fundamentals of Rotational Motion Today’s homework is HW #10, due 9pm, Monday, Apr. 12!!

2. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 2 Announcements Quiz Wednesday, Apr. 9 –At the beginning of the class –Covers 6.7 – what we cover today 3 rd term exam –Monday, Apr. 21, in class –Covers: Ch. 6.7 – what we complete next Wednesday, Apr. 16 –This is the final term exam in the semester –The worst of the three term exams will be dropped from the final grading

3. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 3 Extra-Credit Special Project Derive the formula for the final velocity of two objects which underwent an elastic collision as a function of known quantities m 1, m 2, v 01 and v 02 in page 13 of this lecture note. –20 points extra credit Describe in detail what happens to the final velocities if m1=m2. –5 point extra credit Due: Start of the class this Wednesday, Apr. 9

4. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 4 Linear Momentum The principle of energy conservation can be used to solve problems that are harder to solve just using Newton’s laws. It is used to describe motion of an object or a system of objects. A new concept of linear momentum can also be used to solve physical problems, especially the problems involving collisions of objects. Linear momentum of an object whose mass is m and is moving at a velocity of v is defined as What can you tell from this definition about momentum? What else can use see from the definition? Do you see force? The change of momentum in a given time interval 1.Momentum is a vector quantity. 2.The heavier the object the higher the momentum 3.The higher the velocity the higher the momentum 4.Its unit is kg.m/s

5. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 5 Linear Momentum and Forces What can we learn from this force-momentum relationship? Something else we can do with this relationship. What do you think it is? The relationship can be used to study the case where the mass changes as a function of time. Can you think of a few cases like this? Motion of a meteorite Motion of a rocket The rate of the change of particle’s momentum is the same as the net force exerted on it. When net force is 0, the particle’s linear momentum is constant as a function of time. If a particle is isolated, the particle experiences no net force. Therefore its momentum does not change and is conserved.

6. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 6 Conservation of Linear Momentum in a Two Particle System Consider an isolated system with two particles that do not have any external forces exerting on it. What is the impact of Newton’s 3 rd Law? Now how would the momenta of these particles look like? If particle#1 exerts force on particle #2, there must be another force that the particle #2 exerts on #1 as the reaction force. Both the forces are internal forces, and the net force in the entire SYSTEM is still 0.0. Let say that the particle #1 has momentum p1 p1 and #2 has p2 p2 at some point of time. Using momentum- force relationship And since net force of this system is 0 Therefore The total linear momentum of the system is conserved!!! and

7. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 7 Linear Momentum Conservation Initial Final

8. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 8 More on Conservation of Linear Momentum in a Two Body System What does this mean? As in the case of energy conservation, this means that the total vector sum of all momenta in the system is the same before and after any interactions Mathematically this statement can be written as Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant. From the previous slide we’ve learned that the total momentum of the system is conserved if no external forces are exerted on the system. This can be generalized into conservation of linear momentum in many particle systems.

9. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 9 Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 54-kg woman and one is a 88-kg man. The woman moves away with a speed of +2.5 m/s. Find the recoil velocity of the man. Ex. 6 Ice Skaters No net external force  momentum conserved Solve for V f2

10. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 10 How do we apply momentum conservation? 1.Decide which objects are included in the system. 2.Relative to the system, identify the internal and external forces. 3.Verify that the system is isolated. 4.Set the final momentum of the system equal to its initial momentum. Remember that momentum is a vector.

11. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 11 Collisions Consider a case of a collision between a proton on a helium ion. The collisions of these ions never involve physical contact because the electromagnetic repulsive force between these two become great as they get closer causing a collision. Generalized collisions must cover not only the physical contact but also the collisions without physical contact such as that of electromagnetic ones in a microscopic scale. t F F 12 F 21 Assuming no external forces, the force exerted on particle 1 by particle 2, F 21, changes the momentum of particle 1 by Likewise for particle 2 by particle 1 Using Newton’s 3 rd law we obtain So the momentum change of the system in the collision is 0, and the momentum is conserved

12. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 12 Elastic and Inelastic Collisions Collisions are classified as elastic or inelastic based on whether the kinetic energy is conserved, meaning whether it is the same before and after the collision. A collision in which the total kinetic energy and momentum are the same before and after the collision. Momentum is conserved in any collisions as long as external forces are negligible. Elastic Collision Two types of inelastic collisions:Perfectly inelastic and inelastic Perfectly Inelastic: Two objects stick together after the collision, moving together at a certain velocity. Inelastic: Colliding objects do not stick together after the collision but some kinetic energy is lost. Inelastic Collision A collision in which the total kinetic energy is not the same before and after the collision, but momentum is. Note: Momentum is constant in all collisions but kinetic energy is only in elastic collisions.

13. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 13 Elastic and Perfectly Inelastic Collisions In perfectly Inelastic collisions, the objects stick together after the collision, moving together. Momentum is conserved in this collision, so the final velocity of the stuck system is How about elastic collisions? In elastic collisions, both the momentum and the kinetic energy are conserved. Therefore, the final speeds in an elastic collision can be obtained in terms of initial speeds as From momentum conservation above What happens when the two masses are the same?

14. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 14 The mass of the block of wood is 2.50-kg and the mass of the bullet is 0.0100-kg. The block swings to a maximum height of 0.650 m above the initial position. Find the initial speed of the bullet. Ex. 8 A Ballistic Pendulum What kind of collision? No net external force  momentum conserved Perfectly inelastic collision Solve for V 01 What do we not know?The final speed!! How can we get it?Using the mechanical energy conservation!

15. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 15 Ex. 8 A Ballistic Pendulum, cnt’d Using the solution obtained previously, we obtain Now using the mechanical energy conservation Solve for V f

16. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 16 Two dimensional Collisions In two dimension, one needs to use components of momentum and apply momentum conservation to solve physical problems. m2m2 m1m1 v 1i m1m1 v 1f  m2m2 v 2f  Consider a system of two particle collisions and scatters in two dimension as shown in the picture. (This is the case at fixed target accelerator experiments.) The momentum conservation tells us: And for the elastic collisions, the kinetic energy is conserved: What do you think we can learn from these relationships? x-comp. y-comp.

17. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 17 Example for Two Dimensional Collisions Proton #1 with a speed 3.50x10 5 m/s collides elastically with proton #2 initially at rest. After the collision, proton #1 moves at an angle of 37 o to the horizontal axis and proton #2 deflects at an angle  to the same axis. Find the final speeds of the two protons and the scattering angle of proton #2, . Since both the particles are protons m 1 =m 2 =m p. Using momentum conservation, one obtains m2m2 m1m1 v 1i m1m1 v 1f  m2m2 v 2f  Canceling m p and putting in all known quantities, one obtains From kinetic energy conservation: Solving Eqs. 1-3 equations, one gets Do this at home x-comp. y-comp.

18. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 18 Center of Mass We’ve been solving physical problems treating objects as sizeless points with masses, but in realistic situations objects have shapes with masses distributed throughout the body. Center of mass of a system is the average position of the system’s mass and represents the motion of the system as if all the mass is on that point. Consider a massless rod with two balls attached at either end. The total external force exerted on the system of total mass M causes the center of mass to move at an acceleration given by as if all the mass of the system is concentrated on the center of mass. What does above statement tell you concerning the forces being exerted on the system? m1m1 m2m2 x1x1 x2x2 The position of the center of mass of this system is the mass averaged position of the system x CM CM is closer to the heavier object

19. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 19 Motion of a Diver and the Center of Mass Diver performs a simple dive. The motion of the center of mass follows a parabola since it is a projectile motion. Diver performs a complicated dive. The motion of the center of mass still follows the same parabola since it still is a projectile motion. The motion of the center of mass of the diver is always the same.

20. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 20 Example for Center of Mass Thee people of roughly equivalent mass M on a lightweight (air-filled) banana boat sit along the x axis at positions x 1 =1.0m, x 2 =5.0m, and x 3 =6.0m. Find the position of CM. Using the formula for CM

21. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 21 Velocity of Center of Mass In an isolated system, the total linear momentum does not change, therefore the velocity of the center of mass does not change.

22. Monday, Apr. 7, 2008PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 22 Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 54-kg woman and one is a 88-kg man. The woman moves away with a speed of +2.5 m/s. A Look at the Ice Skater Problem